Upthrust, Archimedes' Principle and Floatation ICSE Class 9: Buoyancy and Relative Density

What Upthrust, Archimedes' Principle and Floatation is about

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Physics. Class IX is examined by your school: one 80-mark paper of 2 hours plus 20 marks of internal assessment. Upthrust, Archimedes' Principle and Floatation is in the 4. Fluids (ii) part of the 2028 syllabus.Source: CISCE ICSE 2028 Physics syllabus, Class IX

Key concepts

Upthrust and its properties

Upthrust is the upward force exerted by a fluid on a body partly or wholly immersed in it.

Why it arises: pressure in a fluid increases with depth, so the upward push on the bottom of the body is greater than the downward push on its top.

Properties of upthrust:

• it acts vertically upwards through the centre of gravity of the displaced fluid (the centre of buoyancy);
• it increases with the volume of the body immersed (up to full immersion);
• it increases with the density of the fluid (more in sea water than in fresh water);
• once the body is fully immersed, it does not depend on depth or on the body's own material.

Worked example

The same iron ball is fully immersed in turn in water (1000 kg m⁻³) and in brine (1200 kg m⁻³). In which is the upthrust greater, and by what factor?

Try it first, then show the solutionSolution

1. Volume displaced is the same in both.
2. Upthrust $=V\rho g$, so it is proportional to the fluid's density.
3. Brine: $\frac{1200}{1000}=1.2$ times greater.

Answer: In brine, 1.2 times

Archimedes' principle and apparent weight

Archimedes' principle: when a body is wholly or partly immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it.

$F_B=V{\rho }_{L}g$, where $V$ is the volume immersed and ${\rho }_L$ the density of the fluid.

Apparent weight $=$ true weight $-$ upthrust $=W-F_B$. The loss in weight in a liquid equals the upthrust.

It can be verified by weighing a body in air and in water with a spring balance and catching the displaced water in an overflow (eureka) can: the loss in weight equals the weight of water collected.

Worked example

A metal ball of volume 100 cm³ and density 7800 kg m⁻³ is fully immersed in water. Find its weight, the upthrust and its apparent weight ($g=10$ m s⁻²).

Try it first, then show the solutionSolution

1. Mass $=7800\times 100\times 10^{-6}=0.78$ kg; $W=7.8$ N.
2. $F_B=100\times 10^{-6}\times 1000\times 10=1$ N.
3. Apparent weight $=7.8-1=6.8$ N.

Answer: 7.8 N; 1 N; 6.8 N

The three cases

Compare the weight $W$ of a fully immersed body with the upthrust $F_B$:

• **Case I: $F_B<W$** (${\rho }_{body}>{\rho }_{fluid}$): the body sinks. In the fluid it has an apparent weight $W-F_B$.
• **Case II: $F_B=W$** (${\rho }_{body}={\rho }_{fluid}$): the body stays where it is placed, fully immersed; apparent weight is zero.
• **Case III: $F_B>W$** (${\rho }_{body}<{\rho }_{fluid}$): the body rises until only part of it is immersed, so that the upthrust on the immersed part equals its weight. It then floats, with apparent weight zero.

Cases II and III are where Archimedes' principle becomes the principle of floatation.

Worked example

A block of volume 0.002 m³ and mass 1.5 kg is released fully under water. What happens? ($g=10$ m s⁻²)

Try it first, then show the solutionSolution

1. $W=15$ N; $F_B$ (fully immersed) $=0.002\times 1000\times 10=20$ N.
2. $F_B>W$: Case III; the block rises.
3. It floats with $\frac{15}{20}=\frac{3}{4}$ of its volume immersed.

Answer: It rises and floats with ¾ of its volume under water.

Relative density

Relative density (RD) of a substance $=\frac{\text{density of the substance}}{\text{density of water at 4 °C}}$. It has no unit. In SI, density $=\text{RD}\times 1000$ kg m⁻³.

RD of a solid denser than water (weigh in air and in water):
$\text{RD}=\frac{\text{weight in air}}{\text{loss of weight in water}}=\frac{W_1}{W_1-W_2}$.

RD of a liquid (weigh a solid in air, in water and in the liquid):
$\text{RD}=\frac{\text{loss of weight in liquid}}{\text{loss of weight in water}}=\frac{W_1-W_3}{W_1-W_2}$.

These work because the loss of weight equals the weight of an equal volume of water (or liquid).

Worked example

A stone weighs 50 N in air, 30 N in water and 34 N in a liquid. Find the RD of the stone and of the liquid.

Try it first, then show the solutionSolution

1. Loss in water $=20$ N; RD of stone $=\frac{50}{20}=2.5$.
2. Loss in liquid $=16$ N; RD of liquid $=\frac{16}{20}=0.8$.

Answer: 2.5 and 0.8

Floatation and the fraction immersed

Principle of floatation: a floating body displaces a weight of fluid equal to its own weight. (Apparent weight of a floating body is zero.)

For a body of total volume $V_1$ and density ${\rho }_1$ floating with volume $V_2$ immersed in a liquid of density ${\rho }_2$:

$V_1{\rho }_{1}g=V_2{\rho }_{2}g$, so $\frac{{\rho }_1}{{\rho }_2}=\frac{V_2}{V_1}$.

The fraction immersed equals the ratio of the densities. A body floats higher in a denser liquid (a ship rises when it sails from a river into the sea).

Worked example

A wooden cube of side 10 cm and density 600 kg m⁻³ floats in water. How deep is it immersed? How deep would it be in oil of density 800 kg m⁻³?

Try it first, then show the solutionSolution

1. Fraction immersed in water $=\frac{600}{1000}=0.6$: depth $=6$ cm.
2. In oil: $\frac{600}{800}=0.75$: depth $=7.5$ cm.

Answer: 6 cm; 7.5 cm

Applications: ship, submarine, iceberg, balloon

• Ship: made of steel (denser than water), but its hollow shape displaces a large volume of water; it sinks until the weight of water displaced equals its own weight. It floats higher in sea water than in river water.
• Submarine: has ballast tanks. Filling them with water increases its weight above the upthrust and it dives; blowing the water out with compressed air makes it lighter and it rises.
• Iceberg: ice (about 917 kg m⁻³) is a little less dense than sea water (about 1030 kg m⁻³), so about $\frac{917}{1030}\approx 0.89$, i.e. nearly nine-tenths, is under water.
• Balloon: filled with a gas lighter than air (hydrogen or helium) or hot air; the upthrust of the air it displaces exceeds its total weight, so it rises.

Worked example

A balloon of volume 10 m³ is filled with hydrogen (density 0.09 kg m⁻³). Air has density 1.3 kg m⁻³ and the empty balloon weighs 50 N. Find the upthrust and the extra load it can lift ($g=10$ m s⁻²).

Try it first, then show the solutionSolution

1. Upthrust $=10\times 1.3\times 10=130$ N.
2. Weight of hydrogen $=10\times 0.09\times 10=9$ N.
3. Load $=130-50-9=71$ N.

Answer: 130 N; 71 N

Physics in real life: Upthrust, Archimedes' Principle and Floatation

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Archimedes' Principle, Upthrust and Apparent Weight
• Floatation, Relative Density and Fraction Immersed

Upthrust, Archimedes' Principle and Floatation formula sheet

Upthrust

$F_B=V{\rho }_{L}g$

V = volume immersed.

Apparent weight

$W_{app}=W-F_B$

Loss of weight = upthrust.

RD definition

$\text{RD}=\frac{{\rho }_{substance}}{{\rho }_{water}}$

No unit; ρ = RD × 1000 kg m⁻³.

RD of a solid

$\text{RD}=\frac{W_1}{W_1-W_2}$

W₁ in air, W₂ in water.

RD of a liquid

$\text{RD}=\frac{W_1-W_3}{W_1-W_2}$

W₃ in the liquid.

Floatation

$\frac{{\rho }_1}{{\rho }_2}=\frac{V_2}{V_1}$

Fraction immersed = density ratio.

Upthrust, Archimedes' Principle and Floatation: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Upthrust, Archimedes' Principle and Floatation

The question patterns to expect, and the method that works for each.

Tricks to remember Upthrust, Archimedes' Principle and Floatation

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Upthrust = displaced weight

  The fluid pushes back with the weight of what you pushed aside.

  Use it when: Archimedes' principle.

• Sink, stay, swim

  Denser: sink. Equal: stay. Lighter: swim (float).

  Use it when: The three cases.

• RD = air over loss

  Solid: weight in air ÷ loss in water. Liquid: loss in liquid ÷ loss in water.

  Use it when: RD numericals.

• Fraction under = density over

  Fraction of a floating body under the liquid = its density ÷ the liquid's density.

  Use it when: Floatation.

Upthrust, Archimedes' Principle and Floatation: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Upthrust, Archimedes' Principle and Floatation

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using the body's density in F_B = Vρg

  Upthrust uses the density of the fluid, not of the body.

  Wrong working

  $F_B=V\times 7800\times g$

  Right working

  $F_B=V\times 1000\times g$ (water)

  ✓Do this: Write ρ_L (liquid) in the formula.

  Usually costs: The answer.

• ✗Mistake: Inverting the RD formula

  RD of a solid is weight in air ÷ loss in water, not loss ÷ weight.

  ✓Do this: Check: a stone must give RD > 1.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Giving RD a unit

  RD is a ratio; it has no unit.

  ✓Do this: Write just the number.

  Usually costs: 1 mark.

• ✗Mistake: Saying a floating body has no weight

  Its apparent weight is zero, but its true weight is balanced by the upthrust.

  ✓Do this: Say 'apparent weight is zero; weight = upthrust'.

  Usually costs: 1 mark.

• ✗Mistake: Using the total volume for a floating body's upthrust

  Only the immersed volume displaces liquid.

  ✓Do this: Use V₂ (immersed) in F_B = V₂ρg.

  Usually costs: The answer.

Practice questions on Upthrust, Archimedes' Principle and Floatation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   A body floats in water with one-fifth of its volume above the surface. Its relative density is:
   (a) 0.2
   (b) 0.8
   (c) 1.25
   (d) 5

   Hint

   Fraction immersed = 4/5.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   Upthrust on a fully immersed body does not depend on:
   (a) the volume of the body
   (b) the density of the liquid
   (c) the density of the body
   (d) g

   Hint

   F_B = Vρ_L g.

   Answer

   (c)

3. Q3 · 2 marks · Short answerExam level

   A solid of volume 200 cm³ is fully immersed in water. Find the upthrust in gf and in newtons ($g=10$ m s⁻²).

   Hint

   Displaced water = 200 g.

   Answer

   200 gf $=0.2\times 10=2$ N.

4. Q4 · 1 mark · Assertion-reasonHarder exam level

   Assertion (A): An egg sinks in fresh water but floats in strong salt water.
   Reason (R): Upthrust increases with the density of the liquid.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Same egg, denser liquid.

   Answer

   (c)

5. Q5 · 3 marks · Short answerHarder exam level

   A metal block weighs 2.7 N in air and 1.7 N in water. Find its volume, density and the apparent weight in a liquid of density 800 kg m⁻³ ($g=10$ m s⁻²).

   Hint

   Upthrust in water = 1 N.

   Answer

   $V=\frac{1}{10^4}=10^{-4}$ m³; density $=\frac{0.27}{10^{-4}}=2700$ kg m⁻³; in liquid: upthrust 0.8 N, apparent weight 1.9 N.

6. Q6 · 3 marks · Short answerHarder exam level

   A log of wood of density 750 kg m⁻³ and volume 0.4 m³ floats in water. Find the volume immersed and the volume above water.

   Hint

   Fraction 0.75.

   Answer

   Immersed $=0.3$ m³; above $=0.1$ m³.

7. Q7 · 3 marks · Short answerHarder exam level

   Describe how you would find the relative density of a liquid using a solid denser than both water and the liquid, and a spring balance.

   Hint

   Three weighings.

   Answer

   Weigh the solid in air ($W_1$), fully in water ($W_2$) and fully in the liquid ($W_3$). RD of liquid $=\frac{W_1-W_3}{W_1-W_2}$, since both losses are weights of the same volume of liquid and of water.

8. Q8 · 3 marks · Short answerHarder exam level

   A piece of cork (density 250 kg m⁻³) of volume 40 cm³ is held fully under water and released. Find the upthrust and the net upward force on it at the moment of release ($g=10$ m s⁻²).

   Hint

   Case III.

   Answer

   Upthrust $=40\times 10^{-6}\times 1000\times 10=0.4$ N; weight $=0.1$ N; net upward force $=0.3$ N.

9. Q9 · 2 marks · Short answerExam level

   Why is it easier to lift a heavy stone under water than in air?

   Hint

   Apparent weight.

   Answer

   Under water the upthrust acts on the stone, so its apparent weight (weight − upthrust) is less than its weight in air.

10. Q10 · 3 marks · Short answerHarder exam level

    A boat has a base area of 6 m². When a group of people get in, it sinks a further 5 cm in fresh water. Find the total mass of the people.

    Hint

    Extra water displaced.

    Answer

    Extra volume $=6\times 0.05=0.3$ m³; mass $=0.3\times 1000=300$ kg.

11. Q11 · 3 marks · Short answerHOTS / competency

    A block of ice floats in a glass of water filled to the brim. Explain, using the principle of floatation, whether the water overflows when the ice melts.

    Hint

    Displaced mass = mass of ice.

    Answer

    The floating ice displaces water whose mass equals its own mass. When it melts, it becomes exactly that mass of water, which fills exactly the volume it was displacing. So the level stays the same: no overflow.

12. Q12 · 4 marks · Long answerHOTS / competency

    A uniform rod floats upright in water with 16 cm of its 20 cm length submerged. How much will be submerged in a liquid of RD 1.25? Will it float in a liquid of RD 0.75?

    Hint

    Density of rod = 0.8 × water.

    Answer

    RD of rod $=\frac{16}{20}=0.8$. In RD 1.25: fraction $=\frac{0.8}{1.25}=0.64$, so 12.8 cm submerged. In RD 0.75: $0.8>0.75$, so it sinks.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define upthrust.Show answer

   The upward force exerted by a fluid on a body immersed in it.

2. 2.State the principle of floatation.Show answer

   A floating body displaces a weight of fluid equal to its own weight.

3. 3.When does a body sink?Show answer

   When its weight is greater than the upthrust (it is denser than the fluid).

4. 4.Formula for RD of a solid by weighing?Show answer

   Weight in air ÷ loss of weight in water.

5. 5.Apparent weight of a floating body?Show answer

   Zero.

6. 6.Why does a ship rise when it enters the sea from a river?Show answer

   Sea water is denser, so less volume is needed to displace its weight.

7. 7.Fraction of an iceberg under water?Show answer

   About nine-tenths.

8. 8.How does a submarine dive?Show answer

   It fills its ballast tanks with water, increasing its weight above the upthrust.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state Archimedes' principle exactly?
• ☐Do you use the fluid's density for upthrust?
• ☐Can you find RD of a solid and of a liquid?
• ☐Can you decide sink, stay or float from densities?
• ☐Can you find the fraction immersed of a floating body?
• ☐Can you explain ships, submarines, icebergs and balloons?
• ☐Do you remember that RD has no unit?

Upthrust, Archimedes' Principle and Floatation: frequently asked questions